Question

hw10 differentiation rules 1 (targets c1, c5; §3.3)
score: 7/9 answered: 7/9
question 8
let (f(x)=14x + 3-7e^{x}). find the equation of the tangent line to the graph of (f(x)) at the point ((0, - 4))
(y=)
question help: video message instructor

Explanation:

Step1: Find the derivative of $f(x)$

The derivative of $14x$ is $14$ (using power - rule $\frac{d}{dx}(ax)=a$), the derivative of the constant $3$ is $0$, and the derivative of $-7e^{x}$ is $-7e^{x}$ (since $\frac{d}{dx}(e^{x}) = e^{x}$). So $f^\prime(x)=14 - 7e^{x}$.

Step2: Evaluate the derivative at $x = 0$

Substitute $x = 0$ into $f^\prime(x)$: $f^\prime(0)=14-7e^{0}=14 - 7\times1=7$. The value of the derivative at $x = 0$ is the slope $m$ of the tangent line.

Step3: Use the point - slope form of a line

The point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(0,-4)$ and $m = 7$. Substituting these values gives $y-(-4)=7(x - 0)$.

Step4: Simplify the equation

$y + 4=7x$, so $y=7x - 4$.

Answer:

$y = 7x-4$